I try to figure the volume of a fillet (rounded angle of radius r) obtain by revolution
I used the shell method and at one point, I need to solve a part of the integral with the form:
$$ V = 2\pi\int_{0}^{r}{\left(u + r\right) \,\sqrt{\,{r^{2} + u^{2}}\,}\,\mathrm{d}u + \mbox{etc}} $$
it's been a while since i haven't dealed with stuff like that, any idea please $?$.
You can make a substitution: $$ \begin{aligned} &\text { substitute } u=r \tan (s) \text { and } d u=r \sec ^{2}(s) d s . \text { Then } \sqrt{r^{2}+u^{2}}=\\ &\sqrt{r^{2} \tan ^{2}(s)+r^{2}}=r \sec (s) \text { and } s=\tan ^{-1}\left(\frac{u}{r}\right)\\ &=2 \pi r \int r \sec ^{3}(s)(r \tan (s)+r) d s \end{aligned} $$
Factor out constants: $=2 \pi r^{2} \int \sec ^{3}(s)(r \tan (s)+r) d s$
Expanding the integrand $\sec ^{3}(s)(r \tan (s)+r)$ gives $r \sec ^{3}(s)+r \tan (s) \sec ^{3}(s):$ $=2 \pi r^{2} \int\left(r \sec ^{3}(s)+r \tan (s) \sec ^{3}(s)\right) d s$
Integrate the sum term by term and factor out constants: $=2 \pi r^{3} \int \tan (s) \sec ^{3}(s) d s+2 \pi r^{3} \int \sec ^{3}(s) d s$
For the integrand $\tan (s) \sec ^{3}(s),$ substitute $p=\sec (s)$ and $d p=\tan (s) \sec (s) d s$ $=2 \pi r^{3} \int p^{2} d p+2 \pi r^{3} \int \sec ^{3}(s) d s$
The integral of $p^{2}$ is $\frac{p^{3}}{3}$ : $=\frac{2}{3} \pi p^{3} r^{3}+2 \pi r^{3} \int \sec ^{3}(s) d s$
Use the reduction formula, $\int \sec ^{m}(s) d s=\dfrac{\sin (s) \sec ^{m-1}(s)}{m-1}+$ $\dfrac{m-2}{m-1} \int \sec ^{-2+m}(s) d s,$ where $m=3:$ $=\frac{2}{3} \pi p^{3} r^{3}+\pi r^{3} \tan (s) \sec (s)+\pi r^{3} \int \sec (s) d s$
The integral of $\sec (s)$ is $\dfrac{\log (\tan (s)+\sec (s))}{G}$ $=\dfrac{2}{3} \pi p^{3} r^{3}+\pi r^{3} \tan (s) \sec (s)+\pi r^{3} \log (\tan (s)+\sec (s))+c o n s t$ STEP 11 Substitute back for $p=\sec (s):$ $$ =\dfrac{2}{3} \pi r^{3} \sec ^{3}(s)+\pi r^{3} \tan (s) \sec (s)+\pi r^{3} \log (\tan (s)+\sec (s))+\text { constant } $$
Substitute back for $s=\tan ^{-1}\left(\dfrac{u}{r}\right)$ $=\dfrac{2}{3} \pi r^{3} \sec \left(\tan ^{-1}\left(\frac{u}{r}\right)\right)^{3}+\pi r^{3} \tan \left(\tan ^{-1}\left(\dfrac{u}{r}\right)\right) \sec \left(\tan ^{-1}\left(\frac{u}{r}\right)\right)+$ $$ \pi r^{3} \log \left(\tan \left(\tan ^{-1}\left(\frac{u}{r}\right)\right)+\sec \left(\tan ^{-1}\left(\frac{u}{r}\right)\right)\right)+\text { constant } $$
Simplify using $\sec \left(\tan ^{-1}(z)\right)=\sqrt{z^{2}+1}$ and $\tan \left(\tan ^{-1}(z)\right)=z:$ $x$ $=\pi r u \sqrt{r^{2}+u^{2}}+\frac{2}{3} \pi\left(r^{2}+u^{2}\right)^{3 / 2}+\pi r^{3} \log \left(\sqrt{\frac{u^{2}}{r^{2}}+1}+\frac{u}{r}\right)+$ consta
Factor the answer a different way: $=\frac{1}{3} \pi \sqrt{r^{2}+u^{2}}\left(2 r^{2}+3 r u+2 u^{2}\right)+\pi r^{3} \log \left(\sqrt{\frac{u^{2}}{r^{2}}+1}+\frac{u}{r}\right)+$ consta
Which is equivalent for restricted $u$ and $r$ values to: Answer: $$ =\frac{1}{3} \pi\left(\sqrt{r^{2}+u^{2}}\left(2 r^{2}+3 r u+2 u^{2}\right)+3 r^{3} \log \left(\sqrt{r^{2}+u^{2}}+u\right)\right)+\text { constant } $$
This is the solution to the indefinite integral, you can plug in your values then